APM346-2015F > Final Exam

FE-6

**Victor Ivrii**:

Consider spherically symmetric solutions of 3D-wave equation

\begin{align}

&&&u_{tt}-u_{rr}-\frac{2}{r} u_r=0, \qquad 0<r<1,\label{6-1}\\

&\text{with boundary conditions } \notag\\

&&&|u(0,t)|<\infty, \qquad u(1,t)=0\label{6-2}\\

&\text{and initial conditions}\notag\\

&&&u(r,0)=1,\qquad u_t(r,0)=0\label{6-3}

\end{align}

and solve by a separation of variables.

Hint. $r R''+2 R'= (rR)''$.

**Vivian Tan**:

$\frac{T''}{T}=\frac{R''}{R}+\frac{2}{r}\frac{R'}{R}=-\omega^2$. One can check that there are only negative eigenvalues, so I have defined it this way. Let's look at the $R$ equation first

$$rR''+2R'=(rR)''=-\omega^2 rR$$

This is a second order ODE in $rR$, with solution $rR=A\cos(\omega r)+B\sin(\omega r)$. However, for $u$ to be bounded as $r\rightarrow 0$, we must have $A=0$. This also gives some insight as to why there were only negative eigenvalues to begin with: out of all possible functions, only $\frac{\sin(r)}{r}$ does not blow up at the origin. The boundary condition implies that $\omega=n\pi$. Therefore $R=B\frac{\sin(n\pi r)}{r}$.

Then solving the $T$ equation gives $T=C\cos(n\pi t)+D\sin(n\pi t)$, but the second boundary condition forces $D$ to be $0$. The general solution is, after absorbing and redefining constants:

$$u(r,t)=\sum_{n=1}^{\infty}A_n\frac{\sin(n\pi r)}{r}\cos(n\pi t)$$

$u(r,0)=\sum_{n=1}^{\infty}A_n\frac{\sin(n\pi r)}{r}=1\Rightarrow\sum_{n=1}^{\infty}A_n\sin(n\pi r)=r\Rightarrow A_n=2\int_{0}^{1}r\sin(n\pi r)dr=\frac{(-1)^{n+1}}{n\pi}$. Finally:

$$u(r,t)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n\pi}\frac{\sin(n\pi r)}{r}\cos(n\pi t)$$

**Bruce Wu**:

Actually $$\lim_{r\rightarrow 0}\frac{\sinh(r)}{r}=1$$

So that does not blow up at the origin either. However the problem with that function (corresponding to positive eigenvalues) is that it can never be $0$ at $r\neq 0$, so the boundary conditions can never be satisfied.

**Vivian Tan**:

--- Quote from: Bruce Wu on December 18, 2015, 11:48:40 PM ---Actually $$\lim_{r\rightarrow 0}\frac{\sinh(r)}{r}=1$$

So that does not blow up at the origin either. However the problem with that function (corresponding to positive eigenvalues) is that it can never be $0$ at $r\neq 0$, so the boundary conditions can never be satisfied.

--- End quote ---

Thank you Bruce Wu, I can't believe I missed that.

**Emily Deibert**:

Also, in case anyone doesn't remember, note that $$\lim_{r \rightarrow 0} \frac{\sin (r)}{r} = 1$$

Navigation

[0] Message Index

[#] Next page

Go to full version